Parsing a string efficiently - r

Question

So I've got a column in my data frame that is essentially one long characteristic string that is used to encode about variables for each record. It might look something like this:
string<-c('001034002025003996','001934002199004888')
But much longer.
The strings are structured so each 6 characters are paired together. So you can look at the string above like this:
001034 002025 003996
001934 002199 004888
The first three characters of these is a code corresponding to a certain variable and the next three correspond to the value of that variable. So the above can be broken down into three columns that look like this:
var001 var002 var003 var004
1 034 025 996 NA
2 934 199 NA 888
I need a way to parse this string and return a data frame with the expanded columns.
I wrote a nested loop that looks like this:
for(i in 1:length(string)){
text <- string[i]
for(j in seq(1,505,6)){
var <- substr(text,j, j+2)
var.value <- substr(text, j+3, j+5)
index <- (as.numeric(var))
df[i, index] <- var.value
}
}
where df is an empty data frame created to receive the data. This works, but is slow on larger amounts of data. Is there a better way to do this?

Answer 1

1) This one-liner produces a character matrix (which can easily be converted to a data.frame if need be). No packages are used.
read.dcf(textConnection(gsub("(...)(...)", "\\1: \\2\n", string)))
giving:
001 002 003 004
[1,] "034" "025" "996" NA
[2,] "934" "199" NA "888"
2) This alternative produces the same matrix. The read.table produces a long form data.frame and then tapply reshapes it to a wide matrix.
long <- read.table(text = gsub("(...)(...)", "\\1 \\2\n", string),
colClasses = "character", col.names = c("id", "var"))
tapply(long$var, list(gl(length(string), nchar(string[1])/6), long$id), c)